Integral and Differential equations. Please Help

[sebosfato]

Anyone here know how to use integral and differential equations? To make graphics, curves and calculate the area under the curve?

Maybe with a good explanation and help i could learn some calculus to be able to calculate somethings for my projects.

For example i would like to be able to create an exel sheet with capacitors charge and discharge curves, and be able to calculate the energy of the capacitor over time calculating the area under the curve.

Anyone can help?

Thanks

[bugler]

Calculations in electricity are not done that way.
Complex numbers are used to do the math which involves just multiplication and division of complex numbers as opposed to integrations and differential equations.

Laplace transformation takes you from differential equations and integrals domain to the simpler complex numbers domain.

You need to read a circuits theory book.

[GraffixWB]

any basic circuit design book will have the equations you are looking for. ODE are more used for RCL circuits. cap discharge time, i think , is just and e^-x type equation and will be clear in a book like basic circuit engineering and analysis by irwin and helms (i am not plugging this book. it was on my shelf as I used it in school). any book will do though.

[sebosfato]

Thanks guys for the answers.. .

I'm wiling to calculate the following..

I thought of discharging a capacitor with big capacitance entirely into a small capacitor. Conserving the charge the voltage of the smaller capacitor must be greater as its energy. The fact is that i want to calculate a discharge of the capacitor into a resistor that would give it a discharge time of 1 second to be able to compare if the energy discharged by the bigger capacitor in one second will be really greater or smaller.

I think that this is the flaw in the maxwells law witch tesla found back 100 years ago.

He clearly stated that he could charge up a capacitor using 10w and than develop 10Mw from the charge in that capacitor.

A transformer can change voltages being by time manipulation or turns ratio, You could charge a 10 times smaller capacitor with the same watts to a voltage 10 times greater with the same time, as 10 times greater capacitor charged to a 10 times smaller voltage. However the smaller capacitor will have 10 times more energy.

So i'm thinking about making a graphic, calculating and comparing the discharge time of two different capacitors in one second with the proper resistor. And Compare if the energy discharged will be grater by calculating the area under the graphic.

I tried to find on electronics books but is not very clear, and i'm not very familiar with exponential equations.

If anyone can help-me please.

Lets take easy values like, 0,1F - 100v - 10R (load)- and 0,01F - 1000v - 100R (load). And the Q will be the same, being 10Q.

Basically i want to know how much power will be dissipated in both cases.

Remembering

Q=V*C
C=Q/V
V=Q/C

Tc=C*R

P=V*I
P=R*I^2



I read that the energy accumulated in capacitors are in joules per second. So If so my theory should hold.

Thanks to all

[SilverToGold]

All your questions can be answered here by just plugging in some numbers.

Capacitor Discharging

Substitute in the initial Voltage Vo and t = 1s to get your final Voltage Vc for your RC time constant of that circuit.
Vc = Vo*e^(-t/RC)

This next equation tells you how much energy is left. Just subtract this from what you started with and you will get the energy dissipated in the 1s time.
E = (C*V^2)/2

.

[sebosfato]

Thanks man, even if it don't calculate the power dissipated on the resistor i will take maybe 25 steps and make an exel sheet to calculate... It will help alot, than will make the calculus by hand.

Thanks again, soon the result.

[sebosfato]

So if i understood well, the theory i developed could be sound, cause you can charge a capacitor lets say with 1ma at 40kv for 25us If we discharge this into a resistor of certain resistance it will discharge within a certain time, given by the product of the resistance and capacitance. So lets say we have a 25nf capacitor charged up to 40kv... This mean that it have Q=V*C we have (2,5*10^-8)*(4*10^4) so we have 1*10^-3 coulombs or 1ma, But happens that if you connect a 1000kohm resistor to this charged capacitor the instant current will be 40Amperes. And the time constant would be 1*10^3*2,5*10^-8 so 2,5*10^-5 seconds or 25us microseconds.

This even being small energy only 1/2*2,5*10^-8*4*10^8 = 5 joules. 1 joule is 1 watt per second so 5 joules we have 5w/s

hum i will keep thinking